# Answer to Question #4189 in Calculus for shane duque

Question #4189

A man 6 ft. tall walks away from a lamppost 16 ft. high at a rate of 5 mi per hr. How fast does the of his shadow move?

Expert's answer

Suppose a man was in point A and the shadow was in point S

As we want to evaluate the speed of the shadow, we need to find:

(S

Angles <OS

OL/(LA+AS

AS

Where H is the height of the lamppost (H = 16ft) and h is the man’s height (h = 6ft).

AS

So, S

Angles <OS

OL/(LA+AB+BS

BS

BS

Then:

(S

= V + V*h/(H - h) = VH/(H - h)

The speed of the shadow is 5*16/(16-6)=8 miles/hour

_{1}at time t = t_{0}. After some time t, the man came to the point B and the shadow – to the point S_{2}. OL is a lamppost.As we want to evaluate the speed of the shadow, we need to find:

(S

_{1}S_{0})/t = (S_{1}B+BS_{2})/tAngles <OS

_{1}L and <CS_{1}A are equal, so:OL/(LA+AS

_{1})=AC/(AS_{1})AS

_{1}*H = AS_{1}*h + LA*hWhere H is the height of the lamppost (H = 16ft) and h is the man’s height (h = 6ft).

AS

_{1 }= LA*h/(H-h)So, S

_{1}B = AB - AS_{1 }= Vt - LA*h/(H-h)Angles <OS

_{2}L and <DS_{2}B are equal, so:OL/(LA+AB+BS

_{2}) = BD/(BS_{2})BS

_{2}*H = LA*h + Vt*h + BS_{2}*hBS

_{2 }= h(LA+Vt)/(H-h)Then:

(S

_{1}S_{2})/t = (S_{1}B+BS_{2})/t = V - LA/t*h/(H - h) + h(LA + Vt)/(H - h)t = V + (h*LA + Vth - LA*h)/((H - h)t)= V + V*h/(H - h) = VH/(H - h)

The speed of the shadow is 5*16/(16-6)=8 miles/hour

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